Question

If f'(x)=-f(x) and f(1)=1, then f(x)=

Answer 1

Don't you know how to solve differential equations like this?

Answer 2

No I do not. I am not good in Calculus.

Answer 3

y'=-y with y(1)=1
Integrate both sides, and insert y(1)=1 to find the integration constant.

Answer 4

Or it should be y'/y = -1 => dy/y = -dx, integrate both sides to get
log(y) = -x+C => y=e^(-x)+C
Now substitute y(1)=1 to find C and complete the solution.

Answer 5

\[
y' = -y \\
\frac{dy}{dx} = -y \\
\frac{1}{y} \frac{dy}{dx} = -1 \\

\int \frac{1}{y} \frac{dy}{dx} dx = \int -1dx \\

\]Let \(u = y\) then \(du = \frac{dy}{dx} dx\) (from chain rule) \[
\int \frac{1}{u}du = -x +C_2 \\
\ln(u) + C_1 = -x + C_2 \\
\ln(y) = -x + C_3 \\
e^{\ln(y)} = e ^{-x + C_3} \\
y = C_4 e^{-x}
\]Now we use \(y(1) = 1\) \[
1 = C_4 e^{-1} \\
e = C_4
\]Plug back into original equation: \[
y = e e^{-x} \\
y = e^{1-x}
\]

Answer 6

Sorry, it should read:
\( log(y)=−x+C=>y=Ce^{−x}\)
Consequently substitute y(1)=1 to find C.